MHB T12.3.11 Find the angle between vectors

[karush]

ok don't know the book answer but think this is ok
suggestions welcome

$\tiny{t12.3.11}\\$
$\textsf{Find the angle between vectors}$
\begin{align*}\displaystyle
u&=\sqrt{3i}-7j\\
v&=\sqrt{3i}+j-2k\\
u \cdot v&=(\sqrt{3})(\sqrt{3}) + (-7)(1)+(0)(-2)\\
&=3-7+0\\
&=-4
\end{align*}
$\textit{next the absolute values}$
\begin{align*}\displaystyle
|u||v|&=\sqrt{(\sqrt{3})^2 +(-7)^2}
\cdot
\sqrt{(\sqrt{3})^2+1^2 + (-2)^2}\\
&=\sqrt{51}\cdot \sqrt{8}\\
&=\sqrt{408}
\end{align*}
$\textit{angle between vectors is}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{-4}{\sqrt{408}} \right]\\
&=1.7701 rad\\
&=101.4^o
\end{align*}

is there online graphing that would demonstrate the 2 vectors and the angle between by inputing the vectors?

 

[skeeter]

$\sqrt{(\sqrt{3})^2+(-7)^2}=\sqrt{52}$ ...

 

[karush]

$\tiny{t12.3.11}\\$
------------corrected-------------------
$\textsf{Find the angle between vectors}$
\begin{align*}\displaystyle
u&=\sqrt{3i}-7j\\
v&=\sqrt{3i}+j-2k\\
u \cdot v&=(\sqrt{3})(\sqrt{3}) + (-7)(1)+(0)(-2)\\
&=3-7+0\\
&=-4
\end{align*}
$\textit{next the absolute values}$
\begin{align*}\displaystyle
|u||v|&=\sqrt{(\sqrt{3})^2 +(-7)^2}
\cdot
\sqrt{(\sqrt{3})^2+1^2 + (-2)^2}\\
&=\sqrt{52}\cdot \sqrt{8}\\
&=\sqrt{416}
\end{align*}
$\textit{angle between vectors is}$
\begin{align*}\displaystyle
\theta&=\cos^{-1}\left[\frac{-1}{\sqrt{26}} \right]\\
&=1.77 rad\\
&=101.4^o
\end{align*}